

A292422


Numbers of the form x = 2^k*p*q, p,q odd primes, such that sigma(x)/phi(x) = 4.


4



418, 3596, 3956, 5396, 8636, 41656, 56536, 393104, 2072608, 2316448, 6543008, 17434528, 135394432, 217023616, 1264918784, 1490909824, 2710540544, 11444858368, 17669583104, 34797058304, 37698861568, 70572901376, 132968907776, 226965472256, 233356030976, 552070776832, 596357220352, 601188468736
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OFFSET

1,1


COMMENTS

A subset of A068390 and A020492 (balanced numbers). Conjectured to be infinite by Broughan and Zhou.
Exactly 130 terms are of the form 2^35 * p * q.
We have phi and sigma are multiplicative and sigma(2^k) / phi(2^k) = 4  2/2^k, and sigma(p)/phi(p) = 1 + 2 / (p1).
So we need (4  2/2^k) * (1 + 2 / (p1)) <= 4 which gives a lower bound on p depending on k; p > nextprime(4*2^k).
We can then, given k and p, solve for q. Without loss of generality, p < q. Then search over the primes and stop for that value of k when p > q.
This method may be refined using insights from the article and/or given some k, solve the system (1 + 2 / (p1)) * (1 + 2 / (q  1)) = (a*m) / (b*m) for p and q where a/b is in lowest terms, m > 0. (End)


LINKS



EXAMPLE

418 = 2*11*19; sigma(418) = 720 = 4*phi(418).


PROG

(PARI) is(n) = my(f = factor(n)); #f~ == 3 && f[2, 2] == 1 && f[3, 2] == 1 && f[1, 1] == 2 && sigma(f) / eulerphi(f) == 4 \\ David A. Corneth, Sep 21 2019


CROSSREFS

Subsequence of A293391 (sigma(x)/phi(x) square).


KEYWORD

nonn


AUTHOR



STATUS

approved



